Question

sin A
Prove that (
1- COSA
COS A
1-cos A
:)
sin A
1-sin A) – 4 CBSE2018- 4M
1-sin A
cos A​

Answer 1

Step-by-step explanation:

LHS-----

$( \frac{ \sin( \alpha ) }{1 - \cos( \alpha ) } - \frac{1 - \cos( \alpha ) }{ \sin( \alpha ) } )( \frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } - \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } ) \\ = ( \frac{ \sin( \alpha )(1 + \cos( \alpha ) ) }{(1 - \cos( \alpha ))(1 + \cos( \alpha )) } - \frac{1 - \cos( \alpha ) }{ \sin( \alpha ) } )( \frac{ \cos( \alpha )(1 + \sin( \alpha )) }{(1 - \sin( \alpha ))(1 + \sin( \alpha ) ) } - \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } ) \\ = ( \frac{ \sin( \alpha )(1 + \cos( \alpha ) ) }{ {1}^{2} - { \cos}^{2} \alpha } - \frac{1 - \cos( \alpha ) }{ \sin( \alpha ) } )( \frac{ \cos( \alpha )(1 + \sin( \alpha )) }{ {1}^{2} - { \sin}^{2} \alpha } - \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } ) \\ =( \frac{ 1 + \cos( \alpha ) }{ \sin( \alpha ) } - \frac{1 - \cos( \alpha ) }{ \sin( \alpha ) } )( \frac{ 1 + \sin( \alpha ) }{ \sin( \alpha ) } - \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } ) \\ =( \frac{2 \cos( \alpha ) }{ \sin( \alpha ) } )( \frac{2 \sin( \alpha ) }{ \cos( \alpha ) } ) \\ = 2 \times 2 \\ = 4$

property used in the solution ----

sin²x+cos²x=1

RHS----

4

Since LHS=RHS

hence prooved